I am looking for an equation form that will give me curves as shown below:
The function should take values of X between 0 and 1, and a shape parameter k with values between 0 and 1, to then yield curves of the form as shown above. Y values should be between 0 and 1 for all values of X and k.
So basically it is like a parabola but with the X value at which the maximum occurs shifting depending on the value of k. I am not too concerned about the slope of the curve towards the endpoints at x = 0 and x = 1.
Any suggestions much appreciated
I think what you want is the Beta distribution $B(x; a, b) =x^a(1-x)^b $ with appropriate normalization.
Since
$\begin{array}\\ B'(x; a, b) &=ax^{a-1}(1-x)^b-x^a(1-x)^{b-1}\\ &=x^{a-1}(1-x)^{b-1}(a(1-x)-bx)\\ &=x^{a-1}(1-x)^{b-1}(a-(a+b)x)\\ \end{array} $
$B(x; a, b)$ has an extreme value at $x_{a, b}=\dfrac{a}{a+b}$ of
$\begin{array}\\ B_{a, b} &=B(x_{a, b}; a, b)\\ &=x_{a, b}^a(1-x_{a, b})^b\\ &=(\dfrac{a}{a+b})^a(1-(\dfrac{a}{a+b}))^b\\ &=(\dfrac{a}{a+b})^a(\dfrac{b}{a+b})^b\\ &=\dfrac{a^ab^b}{(a+b)^{a+b}}\\ \end{array} $
Note that $x_{a, b}=1-x_{b, a}$.
To make the scaled $B(k; a, b)=1$, you want the scaling factor $c(k)$ to satisfy $c(k)B(k; a, b) =c(k)k^a(1-k)^b =1$ so the function is $\dfrac{x^a(1-x)^b}{k^a(1-k)^b} $.